Q114 / 200 · NEET 2012 (Screening) · Physics

A geostationary satellite is orbiting the earth at a height of $5 R$ above that surface of the earth, $R$ being the radius of the earth. The time period of another satellite in hours at a height of $2 R$ from the surface of the earth is

  • A
    5
  • B
    10
  • Correct answer
    C
    $6 \sqrt{2}$
  • D
    $6 / \sqrt{2}$

Solution

From Keplar third's law $T^2 \propto r^3$ Hence, $T_1^2 \propto r_1^3$ So, $\begin{aligned} & T_2^2 \propto r_2^3 \\ & \frac{T_2^2}{T_1^2}=\frac{r_2^3}{r_1^3} \\ &=\frac{(3 R)^3}{(6 R)^3} \end{aligned}$ $\begin{aligned} & \text {or } \frac{T_2^2}{T_1^2}=\frac{1}{8} \\ & T_2^2=\frac{1}{8} T_1^2 \\ & T_2=\frac{24}{2 \sqrt{2}}=6 \sqrt{2} \mathrm{~h} \end{aligned}$

Hence, the correct answer is option: 'C'.

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